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Offline Batroost

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Atmospheric pressure and its effect
« Reply #25 on: 02/05/2007 18:44:22 »
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What do you set the scale upon, to weigh anything, when everything floats in the air, including the scale?

This is a really good fundamental question. If you believ, as Einstien did, in teh equivalence of intertial and gravitational mass then then way to 'weigh' something in a weightless environment is simply to apply a force and see how much the object accelerates. Of course what you are measuring is 'mass' - but this is what most people mean when they say 'weight'.

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Offline Batroost

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« Reply #26 on: 02/05/2007 18:49:54 »
Sorry, just spotted something else:

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I also did say, in the very first message in my theory, “When something falls through the atmosphere, and/or through the water, the fall (or sink) velocity increases at a rate that is globally uniform. It seems like there should be something besides mass/ weight/magnetism that would account for the global uniformity of rates of falling speed. Is it not more logical, that a reliable factor like overhead atmospheric pressure would be the influential factor? Acceleration by overhead and gradually increasing downward (pushing) pressure seems like a far more responsible agent of rising velocity.”

But experiment doesn't support this does it? Sorry to repeat a point, but go back to our discussion on Everest. The air pressure is much less at the top of Everest - there is far less atmosphere above you - but you weight (i.e. acceleration due to gravity) has only reduced by less than 1/3 of one percent. Similarly, the difference in weight from the Poles to the Equator (>0.5%) can't be explained by air pressure as it's possible to measure this value at teh same air pressure. And Air pressure at the surface of the earth changes by +/-10% with no meausreable change in acceleration due to gravity.

How can atmopsheric pressure be responsible for the apparent acceleration due to gravity when it is independent of changes in air pressure?

 

Offline Batroost

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« Reply #27 on: 02/05/2007 18:55:11 »
And there's more...

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If the ship could stop falling and remain motionless in the vacuum by firing a powerful set of retro-engines, do you suppose that everybody would crash down to the floor? Not a chance.

No. The ship, and everything in it, would fall (accelerate) directly to Earth. But the occupants would still feel 'weightless' until the friction with the Earth's atmosphere applied a decelerating (upwards) force. The occupants then found that they were apparently being pushed down into their seats. This is excatly what was reported by the Apollo returnees. But this doesn't support your theory because the deceleration (or 'weight') felt by the astronauts was much greater than their 'weight' at the surface even though they were still in a sealed vessel, and they were high up in the atmosphere.
 

Offline Batroost

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« Reply #28 on: 02/05/2007 19:11:41 »
And here's another thought...

Not all rocket paths are orbital. For example, the early Mercury flights were sub-orbital. One thing they all had in common though was that the occupants in the ship experienced weightlessness as soon as the engines were shut-off. This was regardless of the height within or above the atmopshere that was achieved.

Without exception these sub-orbital flights came back... even those that flew above the atmposhere.

I'm happy that all of these flights fit fine within conventional ideas of mechanics and gravity - how do they fit in with your theory?
 

Offline Batroost

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« Reply #29 on: 02/05/2007 19:17:12 »
Sorry...

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mass is non-scalable in the vacuum

Don't understand this. The concept of a scalable mass is strange to me. A mass is a mass is a mass unless you do something physical to add or remove that mass (or add/remove a very great deal of energy).

Dark Energry/Negative Pressure - but this is only one explanation and as many sources point-out (e.g. http://physicsweb.org/articles/world/17/5/7) the difference between observation and theory here could be 120 order of magnitude! IF negative pressure is real then the effect is going to be so small on any scale with which we are familiar to be completely invisible.


 

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Atmospheric pressure and its effect
« Reply #30 on: 02/05/2007 20:01:01 »
Where you been, Chum?

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(1) Why do things accelerate when dropped in a vacuum e.g. in an evacuated tube or on the moon?

Why wouldn’t it accelerate? Things falling in the vacuum of space travel super-fast because there is no friction, wind resistance or anything. If there was a tall tree on the moon, and you dropped an object from the top of it, it would accelerate immediately and continuously through the vacuum until it hit the moon, would it not? There is only a “skin” of atmosphere there.
Look at http://www.britannica.com/ebi/article-204895

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(2) Why do things accelerate when dropped in a sealed air-filled tube - because from Pascal's law the force on them is equalised?

I believe it would be because you are exerting a pressure on the air that is being pressed down as the object falls to the bottom. There has to be even a slight area around the falling object for the increasingly-compressing air to get forced upward, or the object would get stuck in the tube, so as it falls, the air rushes upwards around the object, leaving a void behind and above the falling object that sucks air downward from the falling object’s motion, just like the race car analogy. Does that sound right?

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Atmospheric pressure and its effect
« Reply #31 on: 03/05/2007 14:37:21 »
Hi Batroost;
Boy, you're wordy, so I guess I have to be, too.

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If you believe, as Einstein did, in the equivalence of intertial and gravitational mass then the way to 'weigh' something in a weightless environment is simply to apply a force and see how much the object accelerates. Of course what you are measuring is 'mass' - but this is what most people mean when they say 'weight'.

But as you know, my theory has no belief that gravity accelerates anything, as gravity is a property of matter. Gravity imparts weight to a mass (volume), and it just falls because it has to. (I hope I didn’t misunderstand why you made the first point, but it doesn’t change my theory. If there’s a way to accurately wheigh something out there, good. I just wasn’t aware of it.)

If most people mean mass when they say “weight”, (maybe including me, sometimes), we are not using the (Oxford) dictionary meaning – “body of matter”, etc. Since a body of matter takes up a volume of space, I would accept the word “volume” here. So if I’m trying to scale the weight of a “volume” in space, my position stands.

I also did say, in the very first message in my theory, “When something falls through the atmosphere, and/or through the water, the fall (or sink) velocity increases at a rate that is globally uniform. It seems like there should be something besides mass/ weight/magnetism that would account for the global uniformity of rates of falling speed. Is it not more logical, that a reliable factor like overhead atmospheric pressure would be the influential factor? Acceleration by overhead and gradually increasing downward (pushing) pressure seems like a far more responsible agent of rising velocity.”

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Air pressure at the surface of the earth changes by +/-10% with no measurable change in acceleration due to gravity.
How can atmospheric pressure be responsible for the apparent acceleration due to gravity when it is independent of changes in air pressure?

Does your own question not make you wonder? 
If the world was absolutely round, instead of bulging at the equator, and if the troposphere, stratosphere, mesosphere, thermosphere, and exosphere were all of uniform circumference around the planet at specific altitudes that never changed, it would all be much simpler, but they are not. The height of the troposphere for example, varies from the equator to the poles. At the equator it’s about 11-12 miles high, at 50°N and 50°S it’s 5½ miles and at the poles just under four miles high. The Troposphere is where we live, and where the atmosphere is the densest.
Acceleration due to “gravity” as it’s called, (which I call, “due to overhead pressure”), is “the same” at sea level around the world. That’s called 1G at sea level. Masses fall at 32 FPS/Sec. from “up in the atmosphere”, down to sea level at a speed of “1G”. Drop it off a high bridge, or from a plane, and it’s still the same maximum speed, and the reason is that the total atmosphere has a total weight of 14.7 lbs.per square inch, on every square inch, at sea level, anywhere around the planet. The vertical column of air pressure above a falling object of any surface area that is dropping straight down, weighs 14.7 per square inch of area at sea level, so the area of the falling object is under the influence of however much atmospheric weight is above it as it begins to fall, increasing above it until it hits sea level, when the area on top of that falling object is the entire 14.7 PSI that normally “rests” upon the area of a size that matches the area of the falling object.
What I’m saying is that it doesn’t matter where it fell from. The rules work the same because the planetary rules of physics governing falling objects have to be able to accommodate every possibility. Big area objects and lightweight objects can “fall victim: to factors like wind resistance that “bend the rules” but if those 2 objects, (such as a wide sheet of steel plate and a feather both fell off a rocket ship, they would fall through the vacuum at exactly the same speed, because there is absolutely nothing there to differentiate between their weight, their volume, their area, or their gravity.
It is therefore logical, that as objects fall from the vacuum into our atmosphere, the “transition rules’ must be of such a nature that they can accommodate anything “coming in”. The total vacuum must have specific absolute rules, and the total atmosphere must have specific absolute rules. The “transition phase” between them has to be a workable set of “transitory rules” that “make sense” in any variation of circumstances. It is the responsibility of science to program the “transitory” tables that allow a computer to plot its way through any circumstance. ‘”Patchwork math is not, and never has been the answer”.

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No. The ship, and everything in it, would fall (accelerate) directly to Earth. But the occupants would still feel 'weightless' until the friction with the Earth's atmosphere applied a decelerating (upwards) force. The occupants then found that they were apparently being pushed down into their seats. This is exactly what was reported by the Apollo returnees. But this doesn't support your theory because the deceleration (or 'weight') felt by the astronauts was much greater than their 'weight' at the surface even though they were still in a sealed vessel, and they were high up in the atmosphere.

You’re saying the same thing as me. I said, “IF the ship could “do that””.
Of course it would fall back into the atmosphere. And yes, it would happen exactly as the Apollo returnees said it did. (They just left the vacuum and “gradually plowed into” molecules of air, so the molecules would make a slow cushion and they would plunge into their seats, thus increasing their “apparent weight”.) Note that the “decelerating (upwards) force was on the ship, and not on the people, so the people pushed downwards as a 3rd Law reaction. The fact that they were still in a “sealed vessel” is redundant.

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Not all rocket paths are orbital. For example, the early Mercury flights were sub-orbital. One thing they all had in common though was that the occupants in the ship experienced weightlessness as soon as the engines were shut-off. This was regardless of the height within or above the atmosphere that was achieved.


Call it sub-orbital, or anything you like. The ship had to be in at least the “fairly-pure“ vacuum for the people to leave the floor. The ship could not sustain an atmosphere for very long if there was ANY atmospheric weight above them. You made the implication that I’m wrong, so (if you don’t mind), please check how long they were in orbit. I don’t know how this could be compared to anything else, but I’ll bet they weren’t up for long.

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Don't understand this. The concept of a scalable mass is strange to me. A mass is a mass is a mass unless you do something physical to add or remove that mass (or add/remove a very great deal of energy).

Please go back to the top of this answer, where we covered “mass, weight, and volume”.

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Dark Energry/Negative Pressure - but this is only one explanation and as many sources point-out (e.g. http://physicsweb.org/articles/world/17/5/7) the difference between observation and theory here could be 120 order of magnitude! IF negative pressure is real then the effect is going to be so small on any scale with which we are familiar to be completely invisible.

Hey! They are still trying to figure out exactly what N.P is, but I’m going with Einstein’s Cosmolgical Constant as did the introductory press article in Nov 2005, about the discovery of N.P. The scale will be small, because NP is a universal “platform” and obviously works at the subatomic level. I’m sticking with what they gave us to work with, not what everybody has not yet figured out about something that nobody can see or measure.

Thanks for your patience.

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Offline Batroost

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Atmospheric pressure and its effect
« Reply #32 on: 04/05/2007 17:56:38 »
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Acceleration due to “gravity” as it’s called, (which I call, “due to overhead pressure”), is “the same” at sea level around the world. That’s called 1G at sea level.

No it isn't - that's precisely the point I'm making when I contrast the acceleartion due to gravity at the Poles with the Equator. Both of the values I've given you are measured at Sea Level. Your 32 fps/s value is only an average value. Gravity is NOT  a constant (look it up!) at Sea Level; nor of course is 14.7 psi anywhere near a constant value either. The difference is that the first varience is spatial, the second chronlogical.

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The ship had to be in at least the “fairly-pure“ vacuum for the people to leave the floor.

No. You get exactly the same effect in any freefalling expermient e.g. the German freefall tower experiments a few years ago. Obviously these experiments are short (limited by the height of the Tower) but any of these experiments disprove immediately that there is any conection between between weightlessness and  vacuum.

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The vertical column of air pressure above a falling object of any surface area that is dropping straight down, weighs 14.7 per square inch of area at sea level, so the area of the falling object is under the influence of however much atmospheric weight is above it as it begins to fall, increasing above it until it hits sea level, when the area on top of that falling object is the entire 14.7 PSI that normally “rests” upon the area of a size that matches the area of the falling object.

Another thought experiment: Take a long clear plastic tube up a monutian. At various places stop, hold the tube vertically and tiem how long it takes something small and dense to fall from one end of the tube to the other. Now repeat the experiment at the same locations but with the top and bottom of the tube capped. Would you expect any difference in teh result? No of course not. But you've already quoted Pascal's law for a sealed vessel (which your tube has now become) as meaning that the pressure exerted on anything within the tube is omni-directional. So the falling of teh obejct clearly has nothing to do with air pressure. (If you're feeling keen you can repeat the experiment after evacuating the tube - the object will fal a little faster...).

Here's another thought - and I think this might be significant? What's the pressure gradient going-up through the atmosphere. I don't know but I'd hazard a guess it doesn't ever exceed 1 bar/100 km or about 1Pa/metre of altitude. So the net difference in Air Pressure between my feet and my head is less than 2 Pa or ~ 0.000147 psi. The area of my head/feet are roughly the same (don't go there!) let's say...20cm x 20cm = 0.04 m2. This means that the net force on me from the pressure difference in atmospeheric height will be of the order of 0.08N, imparting an acceleration of about 0.001 m/s2, or roughly 1/10000 of a g. This is the acceleration that I'd feel due to the weight of the atmopshere both above and below me. Ah. But hold on. That pressure is higher below me than above me, so this is an Upwards acceleration - it's going to slow me down! Of course this is all nonsense because in real freefall the effects of air resistance/drag will quickly dominate but it does illustrate why we don't all feel as if we are walking around with (14.7 psi x area) on our heads!

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But as you know, my theory has no belief that gravity accelerates anything, as gravity is a property of matter. Gravity imparts weight to a mass (volume), and it just falls because it has to.

Just spotted this one... Perhaps this is where I've been mis-understanding some of what you are saying. This quote is pretty much along the lines of General Relativety i.e. Gravity is not 'action at a distance' (as Newton described it), but is a symptom of distortions in space/time brought about by teh presence of mass. A good thumbnail descripton of a geodesic path in space-time might well be that the object is following a 'natural path' rather than the obejct is being accelerated by a force.

Have fun,

Batroost

« Last Edit: 04/05/2007 17:58:21 by Batroost »
 

Offline Bored chemist

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Atmospheric pressure and its effect
« Reply #33 on: 05/05/2007 15:52:29 »
Recent experimental observation of a famous physicist on a plane indicates that freefall leads to zero gravity well inside the earth's atmosphere.
http://www.timesonline.co.uk/tol/news/uk/article1687492.ece

Accurate measurements of mass are done in vacuum chambers to remove the effect of air bouancy.
http://publications.npl.co.uk/npl_web/pdf/cmam88.pdf
Even without any air thimgs still have weight.
 

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Atmospheric pressure and its effect
« Reply #34 on: 06/05/2007 21:35:56 »
Hi Batroost;

I've been on a road trip.

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Acceleration due to “gravity” as it’s called, (which I call, “due to overhead pressure”), is “the same” at sea level around the world. That’s called 1G at sea level.

You said “No it isn't – etc.
When I speak of these things being “the same” globally, I HAVE TO BE speaking in terms of the scientifically accepted averages, of course. The atmosphere is “a living thing”, and there must be averages created for general use. Science has done this, and that’s what I must use. I am creating a theory that differs from those “generally accepted”. I am arguing different effects that might produce different mathematics that derive from the pressure and properties’ “activities”. I can ONLY work with known averages inside which my theory must “perform”. The “new math”, if any had to be created if my theory was found “plausible”, would likely be different from the commonly accepted. I MUST use averages, and I can not accept the typical math tied to the old theories to compute within MY theory.  This general principle works for the practice of my theorizing, and is common across many facets of my theorizing, because “their math” might not be correct if some element in my theory works differently than “theirs”. I am OBLIGATED (by my theoretical variations) to do things this way.


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You get exactly the same effect in any freefalling experiment e.g. the German freefall tower experiments a few years ago. Obviously these experiments are short (limited by the height of the Tower) but any of these experiments disprove immediately that there is any connection between weightlessness and vacuum.

No. You probably get the results you target for when you work near ground level. Again, you’re comparing experiments performed in the troposphere (ground level) range, with things I’m trying to theorize about, at the upper extremes of our atmosphere, which I have simply ascribed as a place where “transition rules” must be occurring. Again, I am OBLIGATED (by my theoretical variations) to do things this way.



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But you've already quoted Pascal's law for a sealed vessel as meaning that the pressure exerted on anything within the tube is omni-directional. So the falling of the object clearly has nothing to do with air pressure.

No. While Pascal’s pressure law does state that a pressure EXERTED in a closed vessel is multi-directional, you can’t extend that to mean that if all you do is cap off the two ends of a tube that it will now contain an (exerting) pressure that is any stronger than the atmospheric pressure that exists outside the tube. All you did was put a tube of glass around an existing volume of atmospheric air. You didn’t claim to have “packed in any extra air” that’s “trying to escape”. Effectively, free air in the closed-ended tube has the same interior “pressure” as a double-open-ended tube held in your other hand.

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Here's another thought – (etc.) Of course this is all nonsense because in real freefall the effects of air resistance/drag will quickly dominate but it does illustrate why we don't all feel as if we are walking around with (14.7 psi x area) on our heads!

No “guessing”, unless you want to promote it as your own theory. I’m theorizing based on logic about attributes of participating factors. We can’t just go making up numbers. I also thank you for not contending (so far) that the vacuum in my head might be one of the big “pitfalls” of getting my ideas across.

Bye.

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fleep

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Atmospheric pressure and its effect
« Reply #35 on: 06/05/2007 21:55:29 »
Hi to Bored Chemist:

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Even without any air thimgs still have weight.

Could I ask you please to go back and read my answer #81752?

Then let's talk.

Thanks

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Offline Batroost

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« Reply #36 on: 06/05/2007 22:54:37 »
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No. While Pascal’s pressure law does state that a pressure EXERTED in a closed vessel is multi-directional, you can’t extend that to mean that if all you do is cap off the two ends of a tube that it will now contain an (exerting) pressure that is any stronger than the atmospheric pressure that exists outside the tube. All you did was put a tube of glass around an existing volume of atmospheric air. You didn’t claim to have “packed in any extra air” that’s “trying to escape”. Effectively, free air in the closed-ended tube has the same interior “pressure” as a double-open-ended tube held in your other hand.

Precisely my point. The fact that the air is enclosed makes no fifference at all to the pressure exerted. So what this proves (reductio ad absurdum) is that whether or not something accelerates in falling has nothing to do with the weight of air above it.

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I can ONLY work with known averages i

Not at all. Work from facts. Working from average values rather than from measured values is a classic scientific error. If you can't make your theory fit measurements then why bother calling it a theory?

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This general principle works for the practice of my theorizing, and is common across many facets of my theorizing, because “their math” might not be correct if some element in my theory works differently than “theirs”. I am OBLIGATED (by my theoretical variations) to do things this way

So is your contention that spatial variations in gravity (i.e. at different points on teh earth's surface) and chronolgical changes in air pressure (same place, different time) are irrelevent to a theory that links air pressure to the acceleration experienced by a falling body? Suggesting that an new mathematical structure migt be needed to explain something is fine - ask Schrodinger! - but this seems to do the reverse, it is claiming something as an explanation where the simplest of maths - that of comparison and correlation - shows that that it isn't true. This isn't a theory because it ignores observation.

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No “guessing”, unless you want to promote it as your own theory. I’m theorizing based on logic about attributes of participating factors. We can’t just go making up numbers

This isn't a guess (estimate would have been a better word to use) nor am I making up numbers - it's what physicists generally refer to as an "order of magnitude calculation". What it clearly shows is that there is no accelerating effect from the atmosphere above you - unless you're prepared to throw away that bit of science relating to gases and pressure and invent something entirely new?

Best wishes,

Batroost





 

Offline Bored chemist

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« Reply #37 on: 07/05/2007 16:26:36 »
"Hi to Bored Chemist:


Quote
Even without any air thimgs still have weight.

Could I ask you please to go back and read my answer #81752?

Then let's talk.

Thanks

fleep"
I did, mainly it talks about spacecraft so far as I can tell it doesn't talk about weighing something in a vacuum chamber. It does talk about sealed containers (like spaceships) full of air in space. It doesn't seem to refer to the fact that if air is the cause of weight and I remove the air how come things still have weight?
If I have missed something please can you clarify it. This is a very simple experiment in principle. A weight on a spring balance in a bell jar would do perfectly well as a model system. If you pull the air out what happens to the weight? I say it falls slightly because there is no longer any air bouying it up. If air pressure is the cause of weight wouldn't it rise as, without air causing the weight, shouldn't the spring contract and pull the weight up?
 

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« Reply #38 on: 09/05/2007 03:18:28 »
Fleep.

It was probably moved because this is the section where you can argue against established thories like you are. As to what you mean BY all messed up i not sure as the moving process shouldnt  remove any content , however if something has gone wrong and some data has been lost we appologise.
 

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Atmospheric pressure and its effect
« Reply #39 on: 09/05/2007 14:43:41 »
To Bored Chemist - Re: messg #84761

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A weight on a spring balance in a bell jar - If you pull the air out what happens to the weight? I say it falls slightly because there is no longer any air buoying it up. If air pressure is the cause of weight wouldn't it rise as, without air causing the weight, shouldn't the spring contract and pull the weight up?

Sorry. I thought I answered this.
Yes, it will go down, and here is why. If you make a vacuum, there is no air to affect the naked activity of the bell jar’s interior, of course. The weight is on a spring scale, but now it’s in a vacuum, so there is no air, so there can be no air pressure inside the jar.
I don’t think I ever said that air pressure affects things in a vacuum, because it can’t. It’s affecting the outside of the bell jar, but not the stuff inside. The weight is imparted by the gravity in the mass that’s on the spring scale, and weight just “falls”, because it IS weight. The result is that if the weight falls as the air is withdrwn, it will go down, not up. A spring is a spring. It will only follow the orders of its tensile properties. The weight is just a "dumbell".
Don’t confuse the bell jar (on the Earth) scenario, with the Shuttle and its contained artificial atmosphere sitting in the vacuum. Stuff falls everywhere if not surrounded by a "globally uncertain" inward pressure as it is in the Shuttle. On Earth, it has to fall because it has weight, even if it’s in a vacuum on Earth. The only ways (I know of) to make it float on the Earth, are to completely surround an object/weight, with either a constant air stream pressure that surrounds it and exceeds the strength of its weight, or if it’s magnetically suspended (e.g. North-North poles), from enough directions to
Make it “float” in the air.

Thanks.

fleep 
 

fleep

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Atmospheric pressure and its effect
« Reply #40 on: 09/05/2007 15:42:02 »
Moderator – (Side note) – Why don’t you put in a field where users can show the messg number to which they are replying? Would this also make it easier to group arguments with different individuals that are corresponding between each other? Let us all know please in a separate message. Thanks.


Now – My reply to Batroost #84491

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The fact that the air is enclosed makes no difference at all to the pressure exerted. So what this proves is that whether or not something accelerates in falling has nothing to do with the weight of air above it.

It proves nothing of the sort. Say a double open-ended glass tube of 1 square inch inside area goes from sea level all the way up to the top of the atmosphere. The atmosphere in the tube is the same as the atmosphere outside it. It weighs 14.7 PSI (average) at sea level in both cases. A ball bearing falls at the same rate in both.
Now we have another tube of the same size and height beside it, except both ends are capped. It’s not a pressurized tube, just full of atmosphere like outside it.

The weight of the atmosphere in both tubes starts off at the top at the atmosphere, and all the air molecules that are “piled” in each tube total 14.7 lbs. per square inch (at average), at the bottom of both tubes. If it started falling from a height where it was almost totally vacuum, the ball bearing would first begin to slow down when it hit “thickening” molecular resistance. Somewhere in that falling process, it has to reach 32 ft./sec. Then it falls faster, until it hits 32 fps/sec. You’re the one who pointed out that the shuttle first gets “cushioned” as it comes in out of space, and I agree. That’s what’s happening with these ball bearings. If the atmosphere “cushions”, it is causing an effect. When that entry effect is finished, (because the atmosphere’s getting thicker and heavier through each cubic inch that the bearings fall), then the atmosphere’s "second" effect takes over.
That downwardly increasing accumulation of molecular weight “cubes” is assigning more and more weight above the progress of the falling object. What happens after the first cushion? The atmospheric “rules of falling weights” takes over – i.e. – 32 fps – 32 fps/sec., (and on down).


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Working from average values rather than from measured values is a classic scientific error. If you can't make your theory fit measurements then why bother calling it a theory?

What do you suppose the scientists are working with when they do globally dispersed experiments that will produce different results? They have to adjust all readings for comparison. They adjust them to the standard that science itself created. It’s called “Datum”, as you know. I’m not doing anything that they are not doing. I’m using their standard.

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So is your contention that spatial variations in gravity (i.e. at different points on the earth's surface) and chronological changes in air pressure (same place, different time) are irrelevant to a theory that links air pressure to the acceleration experienced by a falling body? Suggesting that a new mathematical structure might be needed to explain something is fine - ask Schrodinger! - but this seems to do the reverse, it is claiming something as an explanation where the simplest of maths - that of comparison and correlation - shows that that it isn't true. This isn't a theory because it ignores observation.

My theory does not ignore observation. It is based on the observations that I am explaining.
All right. My theory covers the workings of the planet. If it is correct, it could then be tested using current math methods, and the same measurements would likely result. If I was allowed to pick any single location on the globe, a time, and an altitude from which the object is dropped, what do you suppose science would use to calculate their math, if they didn’t have a “standard average” like 14.7 PSI? Their math uses it as their baseline. Why can’t I?


Quote
This isn't a guess (estimate would have been a better word to use) nor am I making up numbers - it's what physicists generally refer to as an "order of magnitude calculation". What it clearly shows is that there is no accelerating effect from the atmosphere above you - unless you're prepared to throw away that bit of science relating to gases and pressure and invent something entirely new?

How does my above explanation affect your point of view on this, or does it (“sort of”) answer it?

Thanks. Sorry for delay.

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Offline rosy

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Atmospheric pressure and its effect
« Reply #41 on: 09/05/2007 16:08:13 »
Fleep: If you want new features you need to talk to the web-monkey (Daveshorts), ideally via the comments and feedback board. Occasionally magic does happen and new features appear, but it does involve Dave having to reverse engineer the forum software and is very unlikely to happen in the next few weeks as he's pretty well snowed under otherwise.
 

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Atmospheric pressure and its effect
« Reply #42 on: 09/05/2007 16:18:02 »
Moderator – (Side note) – Why don’t you put in a field where users can show the messg number to which they are replying? Would this also make it easier to group arguments with different individuals that are corresponding between each other? Let us all know please in a separate message. Thanks.

You mean like that above?

You just hit the 'quote' button for the message, and it will copy the message to the reply panel, and you can then write your answer underneath it; or if you wish to intersperse your answer with the message you are quoting, then you can simply copy the quote start and quote end tags to have several quoted sections, and your replies in between.

You will note that the quoted text now has a link above it, and if you click on the link, you will be taken back to the original message that you have quoted from.
« Last Edit: 09/05/2007 16:19:55 by another_someone »
 

fleep

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Atmospheric pressure and its effect
« Reply #43 on: 09/05/2007 18:22:41 »
Moderator – (Side note) – Why don’t you put in a field where users can show the messg number to which they are replying? Would this also make it easier to group arguments with different individuals that are corresponding between each other? Let us all know please in a separate message. Thanks.

You mean like that above?

You just hit the 'quote' button for the message, and it will copy the message to the reply panel, and you can then write your answer underneath it; or if you wish to intersperse your answer with the message you are quoting, then you can simply copy the quote start and quote end tags to have several quoted sections, and your replies in between.

You will note that the quoted text now has a link above it, and if you click on the link, you will be taken back to the original message that you have quoted from.

Thank you. "Nuff said".

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Offline Bored chemist

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Atmospheric pressure and its effect
« Reply #44 on: 09/05/2007 19:49:33 »
I know that the pressure varies day by day to the extent of about 10%
I know that the force of gravity does not.

I also know that I can measure the force of gravity at the equator and I can measure the force of gravity at the poles. If I am prepared to sit around waiting for the natural variation in air pressures then I can measure the force of gravity in different places but at the same air pressure. This force varies in exactly the way I would expect with latitude and altitude.
It does not vary ( at any given location) with air pressure.
How is it possible that something that changes (like air pressure) can be responsible for something that stays the same ( like gravity)
Also how come something that is the same (Air pressure when its some particular measured value like 760mmHg) be responsible for something that varies (like gravity at the poles or the equator)
« Last Edit: 09/05/2007 19:55:55 by Bored chemist »
 

Offline Batroost

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Atmospheric pressure and its effect
« Reply #45 on: 09/05/2007 21:33:35 »
Quote
The weight of the atmosphere in both tubes starts off at the top at the atmosphere, and all the air molecules that are “piled” in each tube total 14.7 lbs. per square inch (at average), at the bottom of both tubes

Nope. If a tube is sealed (rigidly) then the atmosphere above the tube does not exert any pressure on the gas within. If this weren't true no-one could survive a trip down to see the Titanic!

Quote
That downwardly increasing accumulation of molecular weight “cubes” is assigning more and more weight above the progress of the falling object. What happens after the first cushion? The atmospheric “rules of falling weights” takes over – i.e. – 32 fps – 32 fps/sec., (and on down).

I can't see how you can argue this when you youself have quoted Pascals gas law in your earlier explanations. The point of the tube 'thought-experiment' is that the falling ball bearing inside  doesn't know whether it has a 100km column of air above it, or a few centimetres and then a rigid cap. It will fall in exactly the same manner. This is consistent with Pascals law - i.e. the gas inside the tube exerting an equal pressure in all directions - and wholly inconsistent with any idea that the air above the ball is somehow pushing it down/accelerating it.

Quote
What do you suppose the scientists are working with when they do globally dispersed experiments that will produce different results? They have to adjust all readings for comparison. They adjust them to the standard that science itself created. It’s called “Datum”, as you know.

This isn't how I was taught to do experiments. Yes, you have to adjust results for datum conditions in order to make meaningful comparisons but this is not the same as only using average values. For example, if I'm interested in the boiling points of different fluids then it makes sense to measure them (with a thermometer) and then adjust separately for measured atmospheric pressure at the time of each experiment. This will give the 'correct' boiling points from which deductions can be made about the relative strengths of binding forces within the liquids. However, if I simply use 'average' air pressure, with no correction, I will get misleading results. Some measurements will be high, some will be low. I will get the wrong answers. Using average values is fine for teaching people principles but it is no way to conduct experiment.

Quote
All right. My theory covers the workings of the planet.

But doesn't seem to explain gravity on teh Moon where there is no atmosphere to speak of? Doesn't the non-universality of your explanation bother you (when compared with more conventional explanations)?

Quote
If it is correct, it could then be tested using current math methods, and the same measurements would likely result. If I was allowed to pick any single location on the globe, a time, and an altitude from which the object is dropped, what do you suppose science would use to calculate their math, if they didn’t have a “standard average” like 14.7 PSI? Their math uses it as their baseline. Why can’t I?

There is nothing magical about 14.7 psi. I often find practical problems where using it would be a mistake. For example, the Reactor Building where I work is an airtight enclosure of, roughly speaking, 90,000 cubic metres. The air pressure within this building has to be adjusted (by cooling/heating systems) so as to be within a few millibar of atmospeheric pressure outside - which is always changing, so some active control of these systems is required. If we don't do this then we hit practical problems such as if someone opens an airlock door a differential pressure of more than a few millibar would (quite literally) blow the door and them away! I'm not joking here - this happened a few years ago in the States where the operators involved did exactly as you suggest and compared a pressure reading within their Reactor Building with "14.7 psi" rather than with the true (lower) air pressure at the time. When the door crashed open it killed one of them and seriously injured another. Science is no different to real-life here. You don't use an average value to discount observations that don't fit a theory. From measurement, air pressure varies where ethe acceleration due to gravity doesn't and vice versa.

No correlation = no theory.

Quote
How does my above explanation affect your point of view on this, or does it (“sort of”) answer it?

No - the more I think about this, the more examples there seem to be that contradict what you suggest.

Sorry,

Best wishes,

Batroost
 

Offline rosy

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« Reply #46 on: 09/05/2007 23:29:38 »
The pressure thing across doors can be a real issue even in a much less effectively sealed building. I work in a chemistry lab, with 22 fume hood extractors running full time and lowering the pressure in the lab. We have active pumps moving air back into the lab, but when these shut down for maintenance several things happen:
- All the flow alarms go off... the pressure indoors is too low for the extrctors to work against effectively.
- Any open windows produce a howling draft, taking anything not nailed down off the worksurface and onto the floor.
- Last time it happened I and the other girls couldn't get the door open against the pressure in the corridor (in my defence my arm was in plaster at the time) and we had to get one of the guys to push it really hard and then hold it until we were ready to let it slam.

No real relevance to topic, but it was quite striking at the time.
 

fleep

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Atmospheric pressure and its effect
« Reply #47 on: 10/05/2007 00:30:47 »
Quote
The weight of the atmosphere in both tubes starts off at the top at the atmosphere, and all the air molecules that are “piled” in each tube total 14.7 lbs. per square inch (at average), at the bottom of both tubes

Nope. If a tube is sealed (rigidly) then the atmosphere above the tube does not exert any pressure on the gas within. If this weren't true no-one could survive a trip down to see the Titanic!

Quote
That downwardly increasing accumulation of molecular weight “cubes” is assigning more and more weight above the progress of the falling object. What happens after the first cushion? The atmospheric “rules of falling weights” takes over – i.e. – 32 fps – 32 fps/sec., (and on down).

I can't see how you can argue this when you youself have quoted Pascals gas law in your earlier explanations. The point of the tube 'thought-experiment' is that the falling ball bearing inside  doesn't know whether it has a 100km column of air above it, or a few centimetres and then a rigid cap. It will fall in exactly the same manner. This is consistent with Pascals law - i.e. the gas inside the tube exerting an equal pressure in all directions - and wholly inconsistent with any idea that the air above the ball is somehow pushing it down/accelerating it.

Quote
What do you suppose the scientists are working with when they do globally dispersed experiments that will produce different results? They have to adjust all readings for comparison. They adjust them to the standard that science itself created. It’s called “Datum”, as you know.

This isn't how I was taught to do experiments. Yes, you have to adjust results for datum conditions in order to make meaningful comparisons but this is not the same as only using average values. For example, if I'm interested in the boiling points of different fluids then it makes sense to measure them (with a thermometer) and then adjust separately for measured atmospheric pressure at the time of each experiment. This will give the 'correct' boiling points from which deductions can be made about the relative strengths of binding forces within the liquids. However, if I simply use 'average' air pressure, with no correction, I will get misleading results. Some measurements will be high, some will be low. I will get the wrong answers. Using average values is fine for teaching people principles but it is no way to conduct experiment.

Quote
All right. My theory covers the workings of the planet.

But doesn't seem to explain gravity on teh Moon where there is no atmosphere to speak of? Doesn't the non-universality of your explanation bother you (when compared with more conventional explanations)?

Quote
If it is correct, it could then be tested using current math methods, and the same measurements would likely result. If I was allowed to pick any single location on the globe, a time, and an altitude from which the object is dropped, what do you suppose science would use to calculate their math, if they didn’t have a “standard average” like 14.7 PSI? Their math uses it as their baseline. Why can’t I?

There is nothing magical about 14.7 psi. I often find practical problems where using it would be a mistake. For example, the Reactor Building where I work is an airtight enclosure of, roughly speaking, 90,000 cubic metres. The air pressure within this building has to be adjusted (by cooling/heating systems) so as to be within a few millibar of atmospeheric pressure outside - which is always changing, so some active control of these systems is required. If we don't do this then we hit practical problems such as if someone opens an airlock door a differential pressure of more than a few millibar would (quite literally) blow the door and them away! I'm not joking here - this happened a few years ago in the States where the operators involved did exactly as you suggest and compared a pressure reading within their Reactor Building with "14.7 psi" rather than with the true (lower) air pressure at the time. When the door crashed open it killed one of them and seriously injured another. Science is no different to real-life here. You don't use an average value to discount observations that don't fit a theory. From measurement, air pressure varies where ethe acceleration due to gravity doesn't and vice versa.

No correlation = no theory.

Quote
How does my above explanation affect your point of view on this, or does it (“sort of”) answer it?

No - the more I think about this, the more examples there seem to be that contradict what you suggest.

Hi Batroost:

Quote
Nope. If a tube is sealed (rigidly) then the atmosphere above the tube does not exert any pressure on the gas within. If this weren't true no-one could survive a trip down to see the Titanic!

Nope, yourself. I said (or meant) that the tube starts off at the “top” of the atmosphere and goes down only to sea level. There is nothing above it that weighs anything, because there’s (virtually) nothing above it. The (non-pressurized) closed tube is full of the weight of the air, which is stratified in the same way as the air outside the closed tube. Heavy air molecules are at the bottom and above that are lighter/thinner air molecules, and at the top is pure helium, then pure hydrogen, then virtually nothing but individual molecules “bleeding” into space.  Every one of the “spheres”, from the troposphere on up, get lighter, and they follow upward successively in lightness and thin-ness, all the way up to where the atmosphere “ends”.

Quote
I can't see how you can argue this when you yourself have quoted Pascals gas law in your earlier explanations. The point of the tube 'thought-experiment' is that the falling ball bearing inside doesn't know whether it has a 100km column of air above it, or a few centimetres and then a rigid cap. It will fall in exactly the same manner. This is consistent with Pascals law - i.e. the gas inside the tube exerting an equal pressure in all directions - and wholly inconsistent with any idea that the air above the ball is somehow pushing it down/accelerating it.
What? You’re still thinking that I’m saying that air “pushes” things down? There is no pressure in the tubes that I’m talking about except atmospheric pressure. The bearings are just falling, and the thickening air weight as they fall increases above them as they drop. Remember the “race-car analogy”? Pascal’s law is about pressures that exceed fluid and/or atmospheric pressure.

Quote
This isn't how I was taught to do experiments. Yes, you have to adjust results for datum conditions in order to make meaningful comparisons but this is not the same as only using average values. For example, if I'm interested in the boiling points of different fluids then it makes sense to measure them (with a thermometer) and then adjust separately for measured atmospheric pressure at the time of each experiment. This will give the 'correct' boiling points from which deductions can be made about the relative strengths of binding forces within the liquids. However, if I simply use 'average' air pressure, with no correction, I will get misleading results. Some measurements will be high, some will be low. I will get the wrong answers. Using average values is fine for teaching people principles but it is no way to conduct experiment.
Come on, Batroost.  Your example is a red herring, and I suspect that you know it. Are you testing me? You’re crossing lines where your analogies are becoming sidetracks.

Quote
But it doesn't seem to explain gravity on the Moon where there is no atmosphere to speak of? Doesn't the non-universality of your explanation bother you (when compared with more conventional explanations)?

O.K. I might have said that “My theory covers the workings of the planet”, but as you know, my entire theory also explains what’s going on inside and outside the shuttle. When I said “the planet”, it referred to the content of that specific message. My theory is for the planet, other atmospheres, non-atmospheres, artificial atmospheres, and the vacuums, both contained in natural atmospheres, and out in the void. I can’t repeat this every time we get into a “local” question.

Quote
There is nothing magical about 14.7 psi. I often find practical problems where using it would be a mistake. For example, the Reactor Building where I work is an airtight enclosure of, roughly speaking, 90,000 cubic metres. The air pressure within this building has to be adjusted (by cooling/heating systems) so as to be within a few millibar of atmospheric pressure outside - which is always changing, so some active control of these systems is required. If we don't do this then we hit practical problems such as if someone opens an airlock door a differential pressure of more than a few millibar would (quite literally) blow the door and them away! I'm not joking here - this happened a few years ago in the States where the operators involved did exactly as you suggest and compared a pressure reading within their Reactor Building with "14.7 psi" rather than with the true (lower) air pressure at the time. When the door crashed open it killed one of them and seriously injured another. Science is no different to real-life here. You don't use an average value to discount observations that don't fit a theory. From measurement, air pressure varies where the acceleration due to gravity doesn't and vice versa. No correlation = no theory.

You’re really reaching, Sir. My theory is what’s going up against the standard use of Datum’s meaning. No particular standard works for every circumstance, but it does for mine. We can cite individually designed systems until the cows come home, but my theory is about gravity in matter in atmospheres and vacuums, etc. That’s it. I can't use any unrelated analogies to propose a different fundamental observation about how things fall.

Quote
The more I think about this, the more examples there seem to be that contradict what you suggest.
Well; I guess I’ll just have to keep trying, won’t I? You, Sir, are a tough sell.

Thanks Batroost.

fleep
 

fleep

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Atmospheric pressure and its effect
« Reply #48 on: 10/05/2007 15:26:02 »
I know that the pressure varies day by day to the extent of about 10%
I know that the force of gravity does not.

I also know that I can measure the force of gravity at the equator and I can measure the force of gravity at the poles. If I am prepared to sit around waiting for the natural variation in air pressures then I can measure the force of gravity in different places but at the same air pressure. This force varies in exactly the way I would expect with latitude and altitude.

It does not vary ( at any given location) with air pressure.
How is it possible that something that changes (like air pressure) can be responsible for something that stays the same ( like gravity)
Also how come something that is the same (Air pressure when its some particular measured value like 760mmHg) be responsible for something that varies (like gravity at the poles or the equator)


Hi Bored Chemist Re: 85701

I (fundamentally) follow Einstein’s view in “General Relativity” that gravitation is not a “force”. If you have followed my theory down, you will know I believe that gravity is what imparts in mass, the “ability” to exhibit weight. Gravity is not a force, and is unable to exert a force outside any atom’s own nucleus. It is not a force in molecular state, mass state, or in anything you want to call the “state”. The “force of gravity”, in my theory, is the influence of the weight of an object. The only things that attract between matter are magnetism and covalence, and even these are selective in their elemental abilities and propensities. I don’t think we would be wrong to say that gravity is weight, but don’t tear off on that tangent with semantic arguments please.

Gravity “stays the same”, as you said it.
You don’t have to sit around anywhere waiting for the “force of gravity” to change. The weight of a mass does not change, (excepting erosion, friction, wear, etc., of course), but different elevations (above sea level for example), around the world will give you different weight readings.

“Air pressure changes”, as you said it.
But the (global) average atmospheric weight at sea level is 14.7 PSI.

(Please now check to confirm that my following simplifications of your two questions above, say exactly what yours said, without using unnecessary example values.)

(Simplifying your first question) -
Q - “How can (changeable) air pressure be responsible for (unchangeable) gravity?”

A - The answer that my theory produces, is that air pressure does not “change” the properties of mass in any way, because air (except as weather, etc.) can not “change” mass.

(Simplifying your second question) -
Q - “How can a specific (measured) air pressure value be responsible for gravity that varies at the poles or the equator?”

A - The answer that my theory produces is the same as my first answer, but, with this qualification:

Regardless of whether we are seeking either a specific (measured) air pressure value, or (making its comparison to) the scientifically created and recognized Datum, (or “global standard value”), the value produced by any specific measurement taken, is always a “product” of that one, (and only that one), location, as measured under the specific air pressure of the moment, as governed by the specific weather conditions existing at that location at the moment.

Thanks B.C.

fleep
 

fleep

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Atmospheric pressure and its effect
« Reply #49 on: 10/05/2007 15:56:47 »
Greetings from Canada:

Fleep here. Re: My ongoing defense about a small part of my entire theory:

In the "earth's atmosphere" part of my theory, everyone seems to keep demanding I must match and explain its workings at every global location’s elevation, weather condition, price of coffee, local hair style, etc. It looks like I’ll be dead long before we get down to the serious consideration that there is a larger fundamental theory here, that everyone is bypassing in an effort to defend traditional scientific “dogma”. If I’m wrong, I will learn that I’m wrong, (and I will learn from that) eventually, but I’m humbly asking anyone to help me finalize this.

I’m happy to get any advice or very pointed questions on what is still unclear. If anyone thinks that they can accept any "part(s)" of my theory as "seemingly logical", could you identify them please? I have no other way to calculate my next directions.  The progression of confusion is growing all the time.

Should I purge all this and state the whole theory again, using the things I have been taught by the questions against which I had to defend my explanations?

My apologies are extended to any who are offended by a frankness that boils over from frustration. I’m getting old here.

Many thanks for all participation.

fleep
 

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